Problem 55 - Differential Equations and Linear Algebra

Question

The alternating symbol \(\epsilon_{i j k}\) is defined by \(\epsilon_{i j k}=\left\\{\begin{aligned} 1, & \text { if }(i j k) \text { is an even permutation of } 1,2,3 \\\\-1, & \text { if }(i j k) \text { is an odd permutation of } 1,2,3 \\ 0, & \text { otherwise. } \end{aligned}\right.\) (a) Write all nonzero \(\epsilon_{i j k},\) for \(1 \leq i \leq 3,1 \leq j \leq 3\) \(1 \leq k \leq 3\) (b) If \(A=\left[a_{i j}\right]\) is a \(3 \times 3\) matrix, verify that $$\operatorname{det}(A)=\sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{i j k} a_{1 i} a_{2 j} a_{3 k}$$.

Step-by-Step Solution

Answer

The nonzero values of the alternating symbol are: \(\epsilon_{123} = 1\), \(\epsilon_{231} = 1\), \(\epsilon_{312} = 1\), \(\epsilon_{132} = -1\), \(\epsilon_{213} = -1\), and \(\epsilon_{321} = -1\). Additionally, the determinant of a 3x3 matrix A can be expressed using the alternating symbol notation as: \(\operatorname{det}(A)=\sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{ijk} a_{1i} a_{2j} a_{3k}\).

Step 1: Find all nonzero values of the alternating symbol

We will start with part (a) of the exercise and find all possible \(\epsilon_{ijk}\) values with the constraint \(1 \leq i \leq 3, 1 \leq j \leq 3, 1 \leq k \leq 3\). It is given that \(\epsilon_{ijk}=1\) for even permutations, and \(\epsilon_{ijk}=-1\) for odd permutations. The permutations and respective values are: Even permutations: $$ (i, j, k) = (1, 2, 3) \quad\Rightarrow\quad \epsilon_{123} = 1 \\ (i, j, k) = (2, 3, 1) \quad\Rightarrow\quad \epsilon_{231} = 1 \\ (i, j, k) = (3, 1, 2) \quad\Rightarrow\quad \epsilon_{312} = 1 $$ Odd permutations: $$ (i, j, k) = (1, 3, 2) \quad\Rightarrow\quad \epsilon_{132} = -1 \\ (i, j, k) = (2, 1, 3) \quad\Rightarrow\quad \epsilon_{213} = -1 \\ (i, j, k) = (3, 2, 1) \quad\Rightarrow\quad \epsilon_{321} = -1 $$ All other permutations will have a value of 0.

Step 2: Prove the given equation for the determinant of a 3x3 matrix

In part (b) of the exercise, we need to prove that the determinant of a 3x3 matrix A can be given by the equation: $$\operatorname{det}(A)=\sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{ijk} a_{1i} a_{2j} a_{3k}$$ Recall that the determinant of a 3x3 matrix is given by: $$\det(A)=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})$$ Let's rewrite this expression using the alternating symbol notation: $$ \operatorname{det}(A) = \epsilon_{123} a_{11} a_{22} a_{33} + \epsilon_{213} a_{12} a_{21} a_{33} + \epsilon_{312} a_{13} a_{21} a_{32} \\ + \epsilon_{132} a_{11} a_{23} a_{32} + \epsilon_{231} a_{12} a_{23} a_{31} + \epsilon_{321} a_{13} a_{22} a_{31} $$ We can see that the summation from the exercise is equal to this expression: $$ \sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{ijk} a_{1i} a_{2j} a_{3k} = \operatorname{det}(A) $$ This shows that the given equation is indeed an alternative expression for the determinant of a 3x3 matrix A.

Key Concepts

Alternating Symbol

Understanding the alternating symbol is crucial in linear algebra, especially when dealing with determinants and tensor calculus. The alternating symbol, denoted as \(\epsilon_{ijk}\), follows a simple set of rules based on the concept of permutations. A permutation refers to an ordered arrangement of objects and, in this context, we are concerned with the order of indices \(i, j, k\).

The alternating symbol takes the value of 1 for even permutations of \(1,2,3\), -1 for odd permutations, and 0 for any other scenarios, including if any two indices are equal. An even permutation can be visualized as a reordering of indices that involves an even number of swaps to return to the original \(1,2,3\) order. Conversely, an odd permutation requires an odd number of swaps.

To improve understanding, remember that the sequence \(1,2,3\) is naturally even. Performing a single swap, say, turning \(1,2,3\) into \(2,1,3\), creates an odd permutation, consequently \(\epsilon_{213}=-1\). Keep swapping and you'll notice that every two swaps will return you to an even permutation.

Permutations in Matrices

Discussing permutations in matrices is essential when addressing more complex problems such as calculating the determinant. In matrices, a permutation involves rearranging the rows or columns. Each permutation can be categorized as 'even' or 'odd' based on the number of swaps needed to achieve the configuration from the original order.

For a 3x3 matrix, you can imagine starting with the rows in their natural order. Any subsequent reordering can be described in terms of permutation of indices \(i, j, k\). By exploring all possible permutations for a 3x3 matrix, we identify that there are six non-zero permutations out of the total of twenty-seven possible ones. For the exercise at hand, understanding this concept enables students to compute the determinant using the alternating symbol formulation.

It's important to see how each term in the determinant corresponds to one of the non-zero permutations, which is why only six terms actually contribute to the determinant's value. This concept is a bridge that helps students transition from a procedural method of calculating determinants to a more conceptual understanding involving the alternating symbol.

Linear Algebra

Linear algebra is a branch of mathematics concentrated on vector spaces and linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces.

The determinant is a fundamental concept within linear algebra, providing a scalar value that can be computed from the elements of a square matrix. The determinant offers insights into the matrix's properties such as singularity, eigenvalues, and more. It is also used for solving systems of linear equations, understanding rotation and scale transformations, and in advanced subjects such as differential equations and quantum mechanics.

In exercises like the given one, linear algebra principles are applied to express complex operations in a compact and theoretically profound way. This use of the alternating symbol to describe the determinant of a 3x3 matrix showcases linear algebra’s capability to simplify and generalize mathematical ideas for further application in complex problems.